[Leetcode] Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

Solution:

1. 非递归

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> inorderTraversal(TreeNode root) {
12         List<Integer> result=new ArrayList<Integer>();
13         if(root==null)
14             return result;
15         Stack<TreeNode> s=new Stack<TreeNode>();
16         s.add(root);
17         TreeNode p=root.left;
18         while(!s.isEmpty()){
19             while(p!=null){
20                 s.add(p);
21                 p=p.left;
22             }
23             TreeNode n=s.pop();
24             result.add(n.val);
25             p=n.right;
26             if(p!=null){
27                 s.add(p);
28                 p=p.left;
29             }
30         }
31         
32         return result;
33     }
34 }

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     
12     ArrayList<Integer> result = new ArrayList<Integer>();
13     
14     public ArrayList<Integer> inorderTraversal(TreeNode root) {
15         
16 
17     Stack<TreeNode> stack = new Stack<TreeNode>();
18         while (root != null || !stack.empty()) {
19             while (root != null) {
20                 stack.push(root);
21                 root = root.left;
22             }
23             if (stack.size() > 0) {
24                 root = stack.pop();
25                 result.add(root.val);
26                 root = root.right;
27             }
28         }
29 
30         return result;
31         
32     }
33 }

2. 递归

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> inorderTraversal(TreeNode root) {
12         List<Integer> result=new ArrayList<Integer>();    
13         myInorderTraversal(root,result);  
14         return result;
15     }
16 
17     private void myInorderTraversal(TreeNode root, List<Integer> result) {
18         // TODO Auto-generated method stub
19         if(root!=null){
20             myInorderTraversal(root.left, result);
21             result.add(root.val);
22             myInorderTraversal(root.right, result);
23         }
24     }
25 }